As has been pointed out, _given_ that you're so obnoxious about otherpeople's English, you should be more careful with yours! The answerto the question as stated is obviously yes, since for example [0,1]is a perfect compact Hausorff space.

The answer to the question you meant to ask is also yes, it seemsto me. Say K is a perfect compact Hausdorff space. EitherK has a connected subset containing more than one point or not.

If C is a subset of K containing p and q, p <> q, then it followsfrom Tietze that there is a continuous f : K -> [0,1] withf(p) = 0 and f(q) = 1; now f(C) must be connected, qed.

On the other hand, if K has no connected subsets withmore than one point: It's easy to construct a continuousmap from K onto the Cantor set, qed. (Say K is the disjointunion of the closed set A_0 and A_1. Map A_0 to theleft half of the Cantor set and A_1 to the right half.Now A_0 is the disjoint union of the closed setsA_00 and A_0,1... repeat countably many timesand you've defined a continuous map onto the Cantor set.)